| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Maximum or minimum velocity |
| Difficulty | Moderate -0.8 This is a straightforward kinematics question requiring basic calculus (integration to find distance, differentiation to find maximum speed) applied to a given velocity function. All steps are routine: substitute t=4 to verify zero speed, integrate v to get displacement, and find the maximum by setting dv/dt=0. The algebraic manipulation is simple and the question structure is highly scaffolded with clear signposting. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v=0\) when it arrives; \(150\,000\!\left(t-\frac{1}{4}t^2\right)=0\Rightarrow t=4\) (on arrival) | B1 | Award for substituting \(t=4\) to obtain \(v=0\). Condone omission of \(t=0\) |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Distance travelled \(s=\int v\,dt\) | M1 | Do not accept multiplication by \(t\) |
| \(s=150\,000\!\left[\frac{1}{2}t^2-\frac{1}{12}t^3\right](+c)\) | A1 | |
| When \(t=4\), \(s=400\,000\) | M1 | Substituting their \(t=4\). Dependent on previous M mark |
| The journey is \(400\,000\) km | A1 | If \(400\,000\) seen award the previous mark |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For maximum speed \(a=\frac{dv}{dt}=0\); \(\frac{dv}{dt}=150\,000\!\left(1-\frac{1}{2}t\right)\Rightarrow t=2\) | B1 | \(t=2\) seen. Accept trial and error method |
| \(v=150\,000\!\left(2-\frac{1}{4}\times2^2\right)=150\,000\); Maximum speed is \(150\,000\ \text{km\,h}^{-1}\) | B1 | CAO |
| [2] |
# Question 5:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v=0$ when it arrives; $150\,000\!\left(t-\frac{1}{4}t^2\right)=0\Rightarrow t=4$ (on arrival) | B1 | Award for substituting $t=4$ to obtain $v=0$. Condone omission of $t=0$ |
| **[1]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distance travelled $s=\int v\,dt$ | M1 | Do not accept multiplication by $t$ |
| $s=150\,000\!\left[\frac{1}{2}t^2-\frac{1}{12}t^3\right](+c)$ | A1 | |
| When $t=4$, $s=400\,000$ | M1 | Substituting their $t=4$. Dependent on previous M mark |
| The journey is $400\,000$ km | A1 | If $400\,000$ seen award the previous mark |
| **[4]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| For maximum speed $a=\frac{dv}{dt}=0$; $\frac{dv}{dt}=150\,000\!\left(1-\frac{1}{2}t\right)\Rightarrow t=2$ | B1 | $t=2$ seen. Accept trial and error method |
| $v=150\,000\!\left(2-\frac{1}{4}\times2^2\right)=150\,000$; Maximum speed is $150\,000\ \text{km\,h}^{-1}$ | B1 | CAO |
| **[2]** | | |5 In a science fiction story a new type of spaceship travels to the moon. The journey takes place along a straight line. The spaceship starts from rest on the earth and arrives at the moon's surface with zero speed. Its speed, $v$ kilometres per hour at time $t$ hours after it has started, is given by
$$v = 37500 \left( 4 t - t ^ { 2 } \right) .$$
(i) Show that the spaceship takes 4 hours to reach the moon.\\
(ii) Find an expression for the distance the spaceship has travelled at time $t$.
Hence find the distance to the moon.\\
(iii) Find the spaceship's greatest speed during the journey.
Section B (36 marks)\\
\hfill \mbox{\textit{OCR MEI M1 2014 Q5 [7]}}